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Tension in a Line

  1. Jan 2, 2008 #1
    [SOLVED] Tension in a Line

    1. The problem statement, all variables and given/known data

    A rope is fixed at both ends on two trees, and a bag is hung in the middle of the rope, causing the rope to sag vertically. If the tree separation is 10 m, the mass of the bag is 5.0 kg, and the sag is 0.2 m, what is the tension in the line?

    2. Relevant equations



    3. The attempt at a solution

    I drew a free body diagram and determine that the angle of the sag was about 5.7 degrees. I'm not to sure where to go from here. Any help would be appreciated.

    Edit: I've attached the free-body diagram. Here I focus on half of the rope which I assume to be 5 m.
     

    Attached Files:

    Last edited: Jan 2, 2008
  2. jcsd
  3. Jan 2, 2008 #2

    Shooting Star

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    I presume that you have taken the angle of sag to be the angle the sagging rope makes with the horizontal. Then tan(theta) = 0.2/5, which gives theta=2.3 deg. (Of course, I can't see the picture yet.)
     
  4. Jan 2, 2008 #3

    HallsofIvy

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    You said the tree separation is 10 m- which has little to do with the length of the rope, but in your picture you show the length of the rope from tree to bag to be 5 m. The 5 meters applies to the horizontal leg of the triangle (the "near" side), not to the hypostenuse. You should have [itex]tan(\theta)= .2/5= 0.04[/itex]. That makes [itex]\theta[/itex] about 2.3 degrees as Shooting star said.

    You know the weight (downward force) on the bag is 5g Newtons and the ropes must offset that. Assume the tension in each rope is T (it would be more interesting if the bag were not in the middle so we would not have symmetry and the tension in the two ropes would be different). The upward force due to the tension in each rope would be [itex]T sin(\theta)[/itex] and so the total upward force would be [itex]2T sin(\theta)= 5g[/itex].
     
    Last edited: Jan 2, 2008
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